Prove that the equation `2x^2+5x y+3y^2+6x+7y+4=0` represents a pair of straight lines. Find the coordinates of their point of intersection and also t

Correct Answer - `(1,-2),tan^(-1)1//5`
The given equation is
`2x^(2)+5xy+3y^(2)+6x+7y+4=0`
Writing (i) as a quadratic equation in x, we get
`2x^(2)+(5y+6)x+3y^(2)+7y+4=0`
`x=(-(5y+6)+-sqrt((5y+6)^(2)-4xx2(3y^(2)+7y+4)))/(4)`
`=(-(5y+6)+-sqrt(25y^(2)+60y+36-24y^(2)-56y-32))/(4)`
`=(-(5y+6)+-sqrt(y^(2)+4y+))/(4)=(-(5+6)+-(y+2))/(4)`
`:.x=(-5y-6+y+2)/(4),(-5y-6-y-2)/(4)`
or `4x+4y+4=0and 4x+6y+8=0`
or `x+y+1=0and 2x+3y+4=0`
Hence , (1) represents a pair of straight lines whose equations are
`x+y+1=0`(1)
and `2x+3y+4=0` (2)
Solving these two equations , the required point of intersection is (1,-2).
Angle between lines ,
`tantheta=|(-1-(-(2)/(3)))/(1+(-1)(-(2)/(3)))|=(1)/(5)*So , theta="tan"^(-1)(1)/(5)`